A mixture of $750\text{kg}$ of alloy of copper and tin contains $25\%$ tin. How much tin must be added so that it becomes $70\%$ of the mixture?

Let's solve this step-by-step by keeping the quantity of copper constant, as only tin is added to the mixture.
Step 1: Find the initial quantity of copper

* Total initial weight of the mixture = $750\text{kg}$
* Initial percentage of tin = $25\%$
* Initial percentage of copper = $100\%-25\%=75\%$

$$\text{Weight of copper}=75\%\text{ of }750=\frac{75}{100}\times 750=562.5\text{kg}$$

Step 2: Use the new mixture conditions
In the final mixture, tin becomes $70\%$. Therefore, copper will be:

$$\text{New percentage of copper}=100\%-70\%=30\%$$

Since no copper was added, the absolute weight of copper remains $562.5\text{kg}$. Let the total weight of the new mixture be $W_{\text{new}}$.

$$30\%\text{ of }{W}_{\text{new}}=562.5$$

$$\frac{30}{100}\times W_{\text{new}}=562.5$$

$$W_{\text{new}}=\frac{562.5\times 100}{30}=1875\text{kg}$$

Step 3: Calculate the amount of tin added

$$\text{Tin added}=W_{\text{new}}-\text{Initial weight}$$

$$\text{Tin added}=1875-750=1125\text{kg}$$